9.4 Combining Two Different Squares

We can fuse two copies of the same square into a larger square (its side being the diagonal of either small square). But Baudhayana presses on to a tougher question:

What if the two squares have different sizes? Can we still merge them into one square whose area equals their total?

His answer, in Sulba-Sutra Verse 1.12, is striking:

Baudhayana’s General Rule The square produced on the diagonal has area equal to the sum of the squares produced on the two sides.

Put plainly: form a right-angled triangle whose two perpendicular sides match the sides of the two squares. The square on its hypotenuse then has area equal to the sum of the two original squares.

Why does it hold? Baudhayana (Verse 2.1) lays out the construction. Set the two squares next to one another. Using a side of the smaller square, mark off a rectangular strip on the larger one, and draw that rectangle’s diagonal, producing a right triangle with perpendicular sides a and b. Now reproduce that triangle four times, positioning them so their hypotenuses outline a slanted four-sided figure.

square on c side b along top, a up the side

Since the four triangles (T, U, X, W) are all congruent, their four hypotenuses are equal, so the slanted figure has four equal sides. Look at each corner, there an angle x from one triangle meets an angle 90^\circ - x from the next, totalling 90^\circ. Four equal sides and four right angles make a square, whose side is the hypotenuse c.

The payoff: the square on the hypotenuse is assembled from exactly the pieces that filled the two original squares (a \times a and b \times b). Therefore

a^2 + b^2 = c^2.

Try This: Combining with Paper Attach two different paper squares of sides a and b in an L-shape. With just two straight cuts, make three pieces, then slide them together into one bigger square, its side is c. Last, build a right triangle with legs a and b, draw the square on its hypotenuse, and confirm that your three pieces blanket it exactly. The two small squares really do “tuck inside” the big one.

Baudhayana and Pythagoras This is the renowned Baudhayana–Pythagoras Theorem. Baudhayana set it down in full generality and essentially modern form in the Sulba-Sutra near 800 BCE, the earliest such statement on record. The Greek thinker Pythagoras (around 500 BCE) likewise prized and studied the result, but lived some two centuries afterwards. The combined name honours both, so that there is no doubt which theorem is meant.

The Theorem Baudhayana–Pythagoras Theorem. If a right-angled triangle has side lengths a, b, and c, with c the hypotenuse (the side opposite the right angle), then a^2 + b^2 = c^2.

Worked Example

A right triangle has shorter sides 6 cm and 8 cm. Find the hypotenuse.

Let a = 6, b = 8. By the theorem, c^2 = a^2 + b^2 = 6^2 + 8^2 = 36 + 64 = 100 \;\Rightarrow\; c = \sqrt{100} = 10\ \text{cm}. Sketch and measure the triangle and the hypotenuse reads about 10 cm, just as the theorem promises.

Worked Example

A ladder reaches 12 m up a wall while its foot stands 9 m out from the base. How long is the ladder?

The wall is vertical and the ground horizontal, so the ladder is the hypotenuse of a right triangle with legs 9 and 12: c^2 = 9^2 + 12^2 = 81 + 144 = 225 \;\Rightarrow\; c = \sqrt{225} = 15\ \text{m}. The ladder is 15 m long.

Figure it Out: Using the Theorem

Practice

  1. A right-angled triangle has shorter sides 5 cm and 12 cm. What is the hypotenuse?
  2. A right-angled triangle has a short side 20 cm and hypotenuse 29 cm. What is the third side?
  3. Using the constructions you have met, how would you build a square whose area is triple that of a given square? Five times the area? (Sulba-Sutra, Verse 1.10.)
  4. With c the hypotenuse, find the missing side:  (i) a=6, b=10  (ii) a=9, b=14  (iii) a=12, c=20  (iv) a=11, b=13  (v) a=2.0, b=4.5.
  1. c^2 = 5^2 + 12^2 = 25 + 144 = 169 \Rightarrow c = \mathbf{13\ \text{cm}}.
  2. b^2 = 29^2 - 20^2 = 841 - 400 = 441 \Rightarrow b = \mathbf{21\ \text{cm}}.
  3. Triple: take a right triangle with legs s (the given side) and s\sqrt2 (the diagonal of the doubled square); its hypotenuse obeys c^2 = s^2 + 2s^2 = 3s^2, so the square on it has area 3s^2. Five times: take legs s and 2s, since s^2 + (2s)^2 = 5s^2; the square on the hypotenuse then has area 5s^2.
    1. c = \sqrt{36+100} = \sqrt{136} \approx \mathbf{11.7}.  (ii) c = \sqrt{81+196} = \sqrt{277} \approx \mathbf{16.6}.  (iii) b = \sqrt{20^2-12^2} = \sqrt{400-144} = \sqrt{256} = \mathbf{16}.  (iv) c = \sqrt{121+169} = \sqrt{290} \approx \mathbf{17.0}.  (v) c = \sqrt{4+20.25} = \sqrt{24.25} \approx \mathbf{4.9}.
Right-Triangle Side Finder

Fill any two of the three boxes (c is the hypotenuse). Leave the unknown blank.